# What type of convolution integral is this?

• I
Convolution has the form

$(f\star g)(t) = \int_{-\infty}^{\infty}f(\tau)g(t-\tau)d\tau$

However, I for my own purposes I have invented a similar but different type of "convolution" which has the form

$(f\star g)(t) = \int_0^{\infty}f(\tau)g(t/\tau)d\tau$

So instead of shifting the function g(t) arithmetically, I shift it multiplicatively with $\tau$ before getting the product and integrating, and I only integrate over the positive domain.

What would be an appropriate name or description of this type of function? Is it discussed somewhere?

haruspex
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Convolution has the form

$(f\star g)(t) = \int_{-\infty}^{\infty}f(\tau)g(t-\tau)d\tau$

However, I for my own purposes I have invented a similar but different type of "convolution" which has the form

$(f\star g)(t) = \int_0^{\infty}f(\tau)g(t/\tau)d\tau$

So instead of shifting the function g(t) arithmetically, I shift it multiplicatively with $\tau$ before getting the product and integrating, and I only integrate over the positive domain.

What would be an appropriate name or description of this type of function? Is it discussed somewhere?
You might call it a geometric (or maybe logarithmic?) convolution, but I'm not sure it isn't effectively covered already.
Suppose we make the following substitutions in the standard form:
ln(μ) for τ
ln(u) for t
xF(x) for f(ln(x)), i.e. f(z)=ezF(ez)
G(x) for g(ln(x)), i.e. g(z)=G(ez)
The RHS becomes ##\int_0^∞F(μ)G(u/μ).dμ##.

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Convolution has the form

$(f\star g)(t) = \int_{-\infty}^{\infty}f(\tau)g(t-\tau)d\tau$

However, I for my own purposes I have invented a similar but different type of "convolution" which has the form

$(f\star g)(t) = \int_0^{\infty}f(\tau)g(t/\tau)d\tau$

So instead of shifting the function g(t) arithmetically, I shift it multiplicatively with $\tau$ before getting the product and integrating, and I only integrate over the positive domain.

What would be an appropriate name or description of this type of function? Is it discussed somewhere?
Where would this be useful? The traditional convolution is extremely fundamental and useful. The set of linear functionals is the same as the set of convolutions with appropriate kernals, g.

Where would this be useful? The traditional convolution is extremely fundamental and useful. The set of linear functionals is the same as the set of convolutions with appropriate kernals, g.

I do not know if my equation is fundamentally interesting but it is a useful operation in my work. If g(x) is some density function for a dataset, and g(x/n) is that function normalized to its mean (g(n) is the mean), then I can essentially add the distribution from g(x) onto another function f(x) as

$\int_0^\infty f(t)g(x/(tn))dt$

So this is a method of multiplying uncertainties described by the distributions in these functions.

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I don't understand that, but if it is working in your application then I have no right to argue with success. I don't recall seeing it before.

haruspex
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If g(x) is some density function for a dataset, and g(x/n) is that function normalized to its mean (g(n) is the mean), then I can essentially add the distribution from g(x) onto another function f(x)
That's not making any sense to me either. Would you mind explaining in more detail?
In what sense are you normalising the distribution?
If μg is the mean of g over 0 to infinity and g(n)=μg then isn't the mean of g(x/n) equal to nμg?

That's not making any sense to me either. Would you mind explaining in more detail?
In what sense are you normalising the distribution?
If μg is the mean of g over 0 to infinity and g(n)=μg then isn't the mean of g(x/n) equal to nμg?
Sorry I think I mean't to write:

$F(x)=\int_0^\infty f(t)g(xn/t)dt$

where $n$ is the mean of $g(x)$:

$n = \int_0^\infty xg(x)dx/\int_0^\infty g(x)dx$

So, one if one had a function $g(xd)$, where $d$ is some constant, the resulting $F(x)$ would be the same for any value of $d$ since the distribution about the mean is the same after normalization.

I found the wikipedia article on this type of convolution and as haruspex said it is called logarithmic convolution: https://en.wikipedia.org/wiki/Logarithmic_convolution

However they write it as

$g*_l f(x) = \int_0^\infty g(x/t)f(t)\frac{dt}{t}$

this is basically what I've written, except that they wrote the derivative as dt/t instead of dt. I don't see how the divisor is necessary though.

mfb
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It is not "necessary", it is a different integral. See it as different weight along the integral. It is useful if you want to consider the ratio of random variables, for example.

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haruspex
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I don't see how the divisor is necessary though.
If you go through what I wrote in post #2 but drop the x factor in xF(x) then I believe you will end up with a dμ/μ in the logarithmic integral. So it is there because it arises naturally. Whether it is appropriate for your problem is another matter, but it does suggest you may have overlooked something.

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